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# M31-22

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14 Jun 2015, 13:15
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For prime numbers $$x$$ and $$y$$, $$x^3*y^5=z^4$$. How many positive factors does positive integer $$z$$ have?

A. One
B. Two
C. Three
D. Four
E. Five
[Reveal] Spoiler: OA

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14 Jun 2015, 13:15
Expert's post
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Official Solution:

For prime numbers $$x$$ and $$y$$, $$x^3*y^5=z^4$$. How many positive factors does positive integer $$z$$ have?

A. One
B. Two
C. Three
D. Four
E. Five

Notice that $$z^4$$ is a perfect square, thus it must have odd number of factors. Since $$z^4$$ equals to $$x^3*y^5$$, then $$x^3*y^5$$ must also have odd number of factors.

Now, if $$x$$ and $$y$$ are different primes, then the number of factors of $$x^3*y^5$$ will be $$(3 + 1)(5 + 1) = 24$$, which is not odd. Therefore $$x$$ and $$y$$ MUST be the same prime.

In this case $$x^3*y^5 = x^3*x^5 = x^8 = z^4$$.

$$x^8 = z^4$$;

$$x^2 = z$$.

Since $$x$$ is prime, then the number of factors of $$z$$ is $$(2 + 1) = 3$$.

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31 Oct 2015, 18:08
"Notice that z4 is a perfect square, thus it must have odd number of factors. Since z4 equals to x3∗y5, then x3∗y5 must also have odd number of factors."

Question about perfect squares. Would z^6 also be a perfect square?

Also, why do x^3 and y^5 need odd factors, given z^4?

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01 Nov 2015, 02:24
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HunterJ wrote:
"Notice that z4 is a perfect square, thus it must have odd number of factors. Since z4 equals to x3∗y5, then x3∗y5 must also have odd number of factors."

Question about perfect squares. Would z^6 also be a perfect square?

Also, why do x^3 and y^5 need odd factors, given z^4?

1. Any integer in even power is a perfect square.
2. $$z^4$$ is a perfect square. A perfect square has odd number of factors. Since $$z^4$$ equals to $$x^3*y^5$$, then $$x^3*y^5$$ (the product of x^3 and y^5, not the multiples separately) must also have odd number of factors.

Finding the Number of Factors of an Integer:

First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.

The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$

Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.

In case of a perfect square, say for $$6^2 = 2^2*3^2$$, the powers of primes will be always even, so when counting the number of factors we'd have $$(even + 1)(even + 1) = (2 + 1)(2 + 1) = 3*3 = odd*odd = odd$$.

1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;

2. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50;

3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);

4. Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: $$36=2^2*3^2$$, powers of prime factors 2 and 3 are even.

Hope it helps.
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06 Nov 2016, 19:46
z cannot be a prefect square unless x=y. x and y are prime numbers, we cant get a prefect square with unequal powers of prime numbers. hence, x=y.

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21 May 2017, 23:41
I think this is a high-quality question and I agree with explanation.

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11 Jun 2017, 12:06
Bunuel wrote:
Official Solution:

For prime numbers $$x$$ and $$y$$, $$x^3*y^5=z^4$$. How many positive factors does positive integer $$z$$ have?

A. One
B. Two
C. Three
D. Four
E. Five

Notice that $$z^4$$ is a perfect square, thus it must have odd number of factors. Since $$z^4$$ equals to $$x^3*y^5$$, then $$x^3*y^5$$ must also have odd number of factors.

Now, if $$x$$ and $$y$$ are different primes, then the number of factors of $$x^3*y^5$$ will be $$(3 + 1)(5 + 1) = 24$$, which is not odd. Therefore $$x$$ and $$y$$ MUST be the same prime.

In this case $$x^3*y^5 = x^3*x^5 = x^8 = z^4$$.

$$x^8 = z^4$$;

$$x^2 = z$$.

Since $$x$$ is prime, then the number of factors of $$z$$ is $$(2 + 1) = 3$$.

SO Bunuel just to reiterate last line. Does that mean the factors are 1,x,z ?
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12 Jun 2017, 04:25
mbsingh wrote:
Bunuel wrote:
Official Solution:

For prime numbers $$x$$ and $$y$$, $$x^3*y^5=z^4$$. How many positive factors does positive integer $$z$$ have?

A. One
B. Two
C. Three
D. Four
E. Five

Notice that $$z^4$$ is a perfect square, thus it must have odd number of factors. Since $$z^4$$ equals to $$x^3*y^5$$, then $$x^3*y^5$$ must also have odd number of factors.

Now, if $$x$$ and $$y$$ are different primes, then the number of factors of $$x^3*y^5$$ will be $$(3 + 1)(5 + 1) = 24$$, which is not odd. Therefore $$x$$ and $$y$$ MUST be the same prime.

In this case $$x^3*y^5 = x^3*x^5 = x^8 = z^4$$.

$$x^8 = z^4$$;

$$x^2 = z$$.

Since $$x$$ is prime, then the number of factors of $$z$$ is $$(2 + 1) = 3$$.

SO Bunuel just to reiterate last line. Does that mean the factors are 1,x,z ?

Absolutely. Say if x = 2, then y = 4 and the factors of 4 are 1, 2, and 4.
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21 Jul 2017, 03:33
Bunuel wrote:
HunterJ wrote:
"Notice that z4 is a perfect square, thus it must have odd number of factors. Since z4 equals to x3∗y5, then x3∗y5 must also have odd number of factors."

Question about perfect squares. Would z^6 also be a perfect square?

Also, why do x^3 and y^5 need odd factors, given z^4?

1. Any integer in even power is a perfect square.
2. $$z^4$$ is a perfect square. A perfect square has odd number of factors. Since $$z^4$$ equals to $$x^3*y^5$$, then $$x^3*y^5$$ (the product of x^3 and y^5, not the multiples separately) must also have odd number of factors.

Finding the Number of Factors of an Integer:

First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.

The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$

Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.

In case of a perfect square, say for $$6^2 = 2^2*3^2$$, the powers of primes will be always even, so when counting the number of factors we'd have $$(even + 1)(even + 1) = (2 + 1)(2 + 1) = 3*3 = odd*odd = odd$$.

1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;

2. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50;

3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);

4. Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: $$36=2^2*3^2$$, powers of prime factors 2 and 3 are even.

Hope it helps.

Hi Bunuel,

Considering the 4th point in the tips for perfect square (great tips btw), we can directly assume in the question that x and y are the same prime numbers since their powers in the question are odd and z is a perfect square? Thanks!
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21 Jul 2017, 03:43
ashikaverma13 wrote:
Bunuel wrote:
HunterJ wrote:
"Notice that z4 is a perfect square, thus it must have odd number of factors. Since z4 equals to x3∗y5, then x3∗y5 must also have odd number of factors."

Question about perfect squares. Would z^6 also be a perfect square?

Also, why do x^3 and y^5 need odd factors, given z^4?

1. Any integer in even power is a perfect square.
2. $$z^4$$ is a perfect square. A perfect square has odd number of factors. Since $$z^4$$ equals to $$x^3*y^5$$, then $$x^3*y^5$$ (the product of x^3 and y^5, not the multiples separately) must also have odd number of factors.

Finding the Number of Factors of an Integer:

First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.

The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$

Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.

In case of a perfect square, say for $$6^2 = 2^2*3^2$$, the powers of primes will be always even, so when counting the number of factors we'd have $$(even + 1)(even + 1) = (2 + 1)(2 + 1) = 3*3 = odd*odd = odd$$.

1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;

2. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50;

3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);

4. Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: $$36=2^2*3^2$$, powers of prime factors 2 and 3 are even.

Hope it helps.

Hi Bunuel,

Considering the 4th point in the tips for perfect square (great tips btw), we can directly assume in the question that x and y are the same prime numbers since their powers in the question are odd and z is a perfect square? Thanks!

_______________
Yes, that's correct.
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26 Jul 2017, 10:20
Just curious multiplication of 2 fractions can be an integer = 2/3 x 3/2 is equal to 1. Is it not possible that x is not equal to y? we find two prime numbers such that forth root of X^3 and Y^5 yield fractions that are reciprocal to each other. Not possible???

P.S. I solved it by thinking that since the question is directly asking how many Prime factor , and because there can be only one correct ans take both X and Y same ( Try to solve with special unique cases when you can)

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24 Aug 2017, 16:19
I think this is a high-quality question and I agree with explanation.

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03 Oct 2017, 04:21
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. I can't think of any positive integer z which satisfies the above equation. the only way according me which satisfies the equation is when x and y are same.
can explain me with a help of an example?

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03 Oct 2017, 04:55
Amanrohra wrote:
For prime numbers $$x$$ and $$y$$, $$x^3*y^5=z^4$$. How many positive factors does positive integer $$z$$ have?

A. One
B. Two
C. Three
D. Four
E. Five

Notice that $$z^4$$ is a perfect square, thus it must have odd number of factors. Since $$z^4$$ equals to $$x^3*y^5$$, then $$x^3*y^5$$ must also have odd number of factors.

Now, if $$x$$ and $$y$$ are different primes, then the number of factors of $$x^3*y^5$$ will be $$(3 + 1)(5 + 1) = 24$$, which is not odd. Therefore $$x$$ and $$y$$ MUST be the same prime.

In this case $$x^3*y^5 = x^3*x^5 = x^8 = z^4$$.

$$x^8 = z^4$$;

$$x^2 = z$$.

Since $$x$$ is prime, then the number of factors of $$z$$ is $$(2 + 1) = 3$$.

I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. I can't think of any positive integer z which satisfies the above equation. the only way according me which satisfies the equation is when x and y are same.
can explain me with a help of an example?

It seems that you did not read the solution carefully.

YES, as it's clearly stated in the solution, x and y MUST be the same prime. For example consider x = y = 2. Note here that unless it is explicitly stated otherwise, different variables CAN represent the same number.
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04 Oct 2017, 01:41
This is a high quality question.

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Re: M31-22   [#permalink] 04 Oct 2017, 01:41
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# M31-22

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